Proposition 2

Let G be a Lie group and h a Lie subalgebra of L(G) admitting a topological supplement. The following conditions are equivalent:

(i) There exist an open subgroup G' of G and a normal Lie subgroup H of G' such that L(H) = h.

(ii) h is an ideal of L(H).

If there exist G' and H with the properties of (i), then L(G') = L(G) and L(H) is an ideal of L(G') by § 3, no. 12, Proposition 47.

Suppose that h is an ideal of L(G). There exists a Lie group F such that L(F) = L(G)/h (Theorem 1). Let h be the canonical morphism of L(G) onto L(F). By Theorem 3 (i), there exist an open subgroup G' of G and a Lie group morphism $\phi$ of G' into F such that $L(\phi) = h$. By § 3, no. 8, the kernel H of $\phi$ is a Lie subgroup of G' and $L(H) = \mathrm{Ker}\ L(\phi) = \mathrm{Ker}\ h = h$. Finally, H is normal in G' since $H = \mathrm{Ker}\ \phi$.